# Schwarzschild geometry for negative masses

Imagine that in a part of our universe there exists negative masses $$M=-|M|<0$$. The metric around this object -- say a black hole -- will be of the form $$ds^2 = -\Big(1+\frac{2|M|}{r}\Big)dt^2 + \Big(1+\frac{2|M|}{r}\Big)^{-1}dr^2+r^2d\Omega^2,$$ where I have put $$G=c=1$$ for simplicity.

From this you can notice that the radial distance $$r= 2|M|$$ has nothing special. At $$r=0$$ there will still be a physical singularity but this time it is not covered by an event horizon because positive masses will easily be able to escape due to repulsion between negative and positive masses.

I expect that light rays will follow the same curves but how about (positive) massive particles? Also, would we be able to see the naked singularity, by using positive particles as a new tool to see?

• There's some relevant material here: en.wikipedia.org/wiki/…
– user4552
Commented Jan 13, 2019 at 15:14
• I expect that light rays will follow the same curves Why would you expect that? Angle of deflection is proportional to mass, change the sign and it will change direction. Commented Jan 13, 2019 at 15:24
• Because photons are massles was my first thought. But indeed, if you look at the deflection formula it will change direction. Commented Jan 13, 2019 at 15:27
• Related: physics.stackexchange.com/q/758783/2451 and links therein. Commented Jun 18 at 7:58

The Schwarzschild metric with negative mass describes an object that repels test objects, whether they have positive or zero mass. The geodesics of those objects, including massless ones, will not be the same as they would have been for an ordinary Schwarzschild black hole. (In the geodesic equation $$\ddot x\sim\Gamma\dot x\dot x$$, the Christoffel symbols $$\Gamma$$ are proportional to $$M$$.) So you would be able to "see" that the negative-mass body is there even with ordinary light, through the way it deflects that light. (And yes, you would also be able to "see" it through its effect on massive test objects.)